1. 3 elements, Mg: 1, S: 1, O: 4
2. 3 elements, Li: 3, P: 1, O: 4
3. 3 elements, H: 2, S: 1, O: 4
1. KOH
2. AlOH
3. AlSO4
Answer:
1.62 g of Al contain the same number of atoms as 6.35 g of cadmium have.
Explanation:
Given data:
mass of cadmium = 6.35 g
Number of atoms of aluminum as 6.35 g cadmium contain = ?
Solution:
Number of moles of cadmium = 6.35 g/ 112.4 g/mol
Number of moles of cadmium = 0.06 mol
Number of atoms of cadmium:
1 mole = 6.022×10²³ atoms of cadmium
0.06 mol × 6.022×10²³ atoms of cadmium/ 1mol
0.36×10²³ atoms of cadmium
Number of atoms of Al:
Number of atoms of Al = 0.36×10²³ atoms
1 mole = 6.022×10²³ atoms
0.36×10²³ atoms × 1 mol /6.022×10²³ atoms
0.06 moles
Mass of aluminum:
Number of moles = mass/molar mass
0.06 mol = m/ 27 g/mol
m = 0.06 mol ×27 g/mol
m = 1.62 g
Thus, 1.62 g of Al contain the same number of atoms as 6.35 g of cadmium have.
The enthalpy of atomization (also atomisation in British spelling) is the enthalpy change that accompanies the total separation of all atoms in a chemical substance (either a chemical element or a chemical compound)
The grams of aluminum that are required to produce 3.5 moles of AlO3 in presence of excess O2 is calculated as below
write the equation for reaction
4 Al + 3O2 =2 Al2O3
by use of mole ratio between Al to Al2O3 which is 4 :2 the moles of Al
=3.5 x4/2 = 7 moles
mass of Al = moles / x molar mass
= 7 moles x27 g/mol =189 grams
N=m/M
n=118/58.93=2
Answer: 2 moles
